User talk:Ohanian

Hello and welcome to Wikipedia! Hope you like it here, and stick around.

Here are some tips to help you get started:

• To sign your posts (on talk pages, for example) use the '~' symbol. To insert just your name, type ~~~ (3 tildes), or, to insert your name and timestamp, use ~~~~ (4 tildes).
• Try the Tutorial, and feel free to experiment in the test area.
• If you need help, post a question at the Help Desk
• Eventually, you might want to read the Manual of Style and Policies and Guidelines.
• Remember Wikipedia:Neutral point of view
• Explore, be bold in editing pages, and, most importantly, have fun!

Good luck!

Meelar (talk) 08:05, Mar 28, 2005 (UTC)

More welcome

Welcome also to Wikipedia:WikiProject Mathematics. This page has good math resources, and a talk page where math-related topics are discussed. If you have not already done so, you can sign in your name at the list of participants.

Thank you for all your work to square root. Oleg Alexandrov 02:52, 15 May 2005 (UTC)

$\mathbf {Y} =\mathbf {K} \mathbf {P}$

we multiply both side by the Transpose of K ie. $\mathbf {K} ^{\operatorname {T} }$

$\mathbf {K} ^{\operatorname {T} }\mathbf {Y} =\mathbf {K} ^{\operatorname {T} }\mathbf {K} \mathbf {P}$

we can put a bracket on $\mathbf {K} ^{\operatorname {T} }\mathbf {K}$

$\mathbf {K} ^{\operatorname {T} }\mathbf {Y} =\left(\mathbf {K} ^{\operatorname {T} }\mathbf {K} \right)^{}\mathbf {P}$

Multiply both sides by the inverse of $\left(\mathbf {K} ^{\operatorname {T} }\mathbf {K} \right)^{}$

$\left(\mathbf {K} ^{\operatorname {T} }\mathbf {K} \right)^{-1}\mathbf {K} ^{\operatorname {T} }\mathbf {Y} =\left(\mathbf {K} ^{\operatorname {T} }\mathbf {K} \right)^{-1}\left(\mathbf {K} ^{\operatorname {T} }\mathbf {K} \right)^{}\mathbf {P}$

The term $\left(\mathbf {K} ^{\operatorname {T} }\mathbf {K} \right)^{-1}\left(\mathbf {K} ^{\operatorname {T} }\mathbf {K} \right)^{}$ on the right hand side cancels out

$\left(\mathbf {K} ^{\operatorname {T} }\mathbf {K} \right)^{-1}\mathbf {K} ^{\operatorname {T} }\mathbf {Y} =\mathbf {P}$

Rearrange

$\mathbf {P} =\left(\mathbf {K} ^{\operatorname {T} }\mathbf {K} \right)^{-1}\mathbf {K} ^{\operatorname {T} }\mathbf {Y}$

$\mathbf {Y} =\mathbf {K} \mathbf {P}$

we multiply both side by the Inverse of K ie. $\mathbf {K} ^{-1}$

$\mathbf {K} ^{-1}\mathbf {Y} =\mathbf {K} ^{-1}\mathbf {K} \mathbf {P}$

The term $\mathbf {K} ^{-1}\mathbf {K}$ on the right hand side cancels out

$\mathbf {K} ^{-1}\mathbf {Y} =\mathbf {P}$

rearrange

$\mathbf {P} =\mathbf {K} ^{-1}\mathbf {Y}$

Definition

Suppose g : Rⁿ → R^m is a function such that each of its first-order partial derivatives exist on Rⁿ. This function takes a point p ∈ Rⁿ as input and produces the vector g(p) ∈ R^m as output. Then the Jacobian matrix of g is defined to be an m×n matrix, denoted by J, whose (i,j)th entry is ${\textstyle \mathbf {J} _{ij}={\frac {\partial g_{i}}{\partial p_{j}}}}$, or explicitly $$\mathbf {J} ={\begin{bmatrix}{\dfrac {\partial \mathbf {g} }{\partial p_{1}}}&\cdots &{\dfrac {\partial \mathbf {g} }{\partial p_{n}}}\end{bmatrix}}={\begin{bmatrix}\nabla ^{\mathrm {T} }g_{1}\\\vdots \\\nabla ^{\mathrm {T} }g_{m}\end{bmatrix}}={\begin{bmatrix}{\dfrac {\partial g_{1}}{\partial p_{1}}}&\cdots &{\dfrac {\partial g_{1}}{\partial p_{n}}}\\\vdots &\ddots &\vdots \\{\dfrac {\partial g_{m}}{\partial p_{1}}}&\cdots &{\dfrac {\partial g_{m}}{\partial p_{n}}}\end{bmatrix}}$$ where $\nabla ^{\mathrm {T} }g_{i}$ is the transpose (row vector) of the gradient of the $i$-th component.

$$\mathbf {p} ={\begin{bmatrix}A\\B\\C\\\end{bmatrix}}$$

Mu

A correction: if a theist asks an atheist "Do you believe in the ideas of atheism" and the atheist says "no", the atheist is still an atheist (in the general sense) as long as he doesn't believe in the ideas of theism; if he says "no" he probably means that he's not a specific type of atheist, but he's still a nontheist in general as long as he doesn't believe in theism. Additionally, many atheists do consider explicit atheism a belief of some sort (even though it's also the rejection of a belief), so those ones would be perfectly fine answering "yes". There are better uses of the answer "mu" to questions (or "not"). And, of course, such a joke doesn't belong on the article page, being so non-noteworthy. Though thanks for sharing. -Silence 15:54, 5 October 2005 (UTC)

Ali Sina

Please see User_talk:Irishpunktom#Ali_Sina, where I explain why it can not be included in the article. The problem is that it is original research and can not be included as such. (In your edit summary you say that it's "evidence". But why is it evidence and according to who, except you and your research?) -- Karl Meier 09:00, 14 October 2005 (UTC)

Tim Priest

Thanks for submitting your article on Tim Priest, but Wikipedia cannot host copyrighted material, and so your article was deleted. The page you submitted was copied from here. If you would like to create an article on Priest without using the copyrighted content, then you are more than welcome to do so. --bainer (talk) 10:32, 14 December 2005 (UTC)

Thank you ;) Remember in future that all contributions must not violate any copyright. --bainer (talk) 10:37, 14 December 2005 (UTC)

calculating numerical value for binary logarithmn

could you please answer the questions on the talk page there? you removed a speedy delete notice, and the notice specifically says not to. I don't beleive you've satisfactorily asserted the why it should be kept . . . --He:ah? 07:03, 8 April 2006 (UTC)

Okay, well, uh, could i just request that you get some of the math people here to look at it, like find someone from the wiki-project or something? because it needs some clean-up- categories, an intro paragraph sort of explaining the what the heck it is, stuff like that. Especially a category. And that article is way out of my league when it comes to editing and whatnot . . . So just keep up the good work, try and get some help with turning it into a proper wikipedia article, and happy editing . . . --He:ah? 08:50, 10 April 2006 (UTC)

Numerical value of a Logarithm

A while back, you posted python code to find the logarithm of any value with any base at Logarithm. Unfortunatly, some of us (like me), don't understand python and aren't willing to learn, so we can't understand it. Could you please write why and how this program works (e.g., its algorithm).

Thank you, Lee S. Svoboda tɑk 19:24, 21 May 2006 (UTC)

Mostly unrelated: If you wanted something to put on your user page, you can put on a python user box

Sure which do you prefer? Pseudocode or C? Here it is in pseudocode

#
# Function in pseudocode
#
function log(float N,float X)
{  epsilon = 0.000000000001
   integer_value=0
   while (X < 1)
   {
      integer_value = integer_value - 1
      X = X * N
   }
   while (X >= N)
   {
      integer_value = integer_value + 1
      X = X / N
   }
   decfrac = 0.0
   partial = 0.5
   X=X*X
   while (partial > epsilon)
   {
      while (X >= N)
      {
         decfrac = decfrac + partial
         X = X / N
      }
      partial = partial / 2
      X=X*X
   }
   return (integer_value + decfrac)
}

Ohanian 01:15, 22 May 2006 (UTC)

Reference desk etiquette

In the post made by Kate about her hearing angels in dreams, other respondents were trying to suggest she see a psychiatrist in the most gentle and sympathetic manner, and I felt good about being among such company in WP - until I read your cruel response - of course she's ill - and you laugh at her? Adambrowne666 02:57, 26 August 2006 (UTC)

Request

Is it possible for me to receive HQ version of Cutie Honey cover you took picture of? I was just wondering. It is ok to refuse, and thanks for any input you'll put in. Dooly00000 (talk · contribs · count)

I'm sorry but I don't have the original anymore. Ohanian 12:56, 2 December 2006 (UTC)

Maths Competition

I'm thinking about starting a maths competition. One which any one can participate and any methods can be used in the competition including astrology, palmistry or numerology.

The objective is to estimate the true probability of obtaining "Head" on a bias coin. The true probability of obtain head with the bias coin can be expected range from 0.1 to 0.9 , in fact the coin toss is simulated by using a table of random numbers.

The procedure of the competition is as follows. The outcome of the coin toss would be annouced and each competitor would display their best estimate using all the knowledge they have obtained.

The full competition would ran for 40 rounds (thus involving 40 coin toss).

The winner would be the participant which has the lowest sum of error square.

A dummy run (of 3 rounds) is as follows: Assume that there are 3 competitors A, B and C.

The judge calls out "Head" and A displays "1.0", B displays "0.6", and C displays "0.5".

The judge calls out "Tail" and A displays "0.0", B displays "0.4", and C displays "0.4".

The judge calls out "Head" and A displays "0.7", B displays "0.7", and C displays "0.5".

The competition has ended and the judge announces the true probability of "Head" for the coin is 0.55 then the scores are calculated.

Coin Toss	Actual value	A's estimate	A's error square	B's estimate	B's error square	C's estimate	C's error square
Head	0.55	1.0	0.2025	0.6	0.0025	0.5	0.0025
Tail	0.55	0.0	0.3025	0.4	0.0225	0.4	0.0225
Head	0.55	0.7	0.0225	0.7	0.0225	0.5	0.0025
		Total	0.5275	Total	0.0475	Total	0.0275

The judge declares C the winner for having the smallest total of error square.

My question is this. If you want to win the competition, what is the best strategy and method for calculating the probability of "Head" in order to obtain the lowest possible error square score. Ohanian 01:31, 29 April 2007 (UTC)

0.999...

Ohanian saids " I have a wonderfully elegant proof that .999... = 1 but this margin is too short for me to write it down, however I shall not be a Fermat person and WILL WRITE IT IN THIS DAMNED MARGIN anyway! "

Let $U={\mbox{unity}}=1\,$

Let $N={\mbox{nines}}=0.999...\,$

Proof by contradiction.

Assume that $N\neq U$. This means that there are only two cases that can follow.

CASE 1 : $N>U$

$N>U\,$

$\downarrow$

$\sum _{k=1}^{\infty }{\frac {9}{10^{k}}}>1$

$\downarrow$

${\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+...>1\,$

$\downarrow$

Contradiction!!!

CASE 2 : $N<U$

$N<U\,$

$\downarrow$

$U+N<U+U$

$\downarrow$

$U+N<2U$

$\downarrow$

$(U+N)\cdot (U-N)<2U\cdot (U-N)$

$\downarrow$

$U^{2}-N^{2}<2U\cdot (1-\sum _{k=1}^{\infty }{\frac {9}{10^{k}}})$

$\downarrow$

$U^{2}-N^{2}<2U\cdot (\lim _{k\to \infty }{\frac {1}{10^{k}}})$

$\downarrow$

$U^{2}-N^{2}<2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$U^{2}<N^{2}+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$U^{2}<N^{2}+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}<N+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$U^{2}<N+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$U<N+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

${\mbox{ }}$

${\frac {U}{2}}<{\frac {N}{2}}+{\frac {2}{2}}\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$0.5<0.499...995+0.000...001$

$\downarrow$

Contradiction!!!

Now since both CASE 1 and CASE 2 results in contradiction, the only conclusion we can come up with is that 0.999... = 1

I rest my case. Ohanian 01:01, 5 May 2007 (UTC)

Fair use rationale for Image:Cutey honey small.jpg

Thanks for uploading or contributing to Image:Cutey honey small.jpg. I notice the image page specifies that the image is being used under fair use but there is not a suitable explanation or rationale as to why each specific use in Wikipedia constitutes fair use. Please go to the image description page and edit it to include a fair use rationale.

If you have uploaded other fair use media, consider checking that you have specified the fair use rationale on those pages too. You can find a list of 'image' pages you have edited by clicking on the "my contributions" link (it is located at the very top of any Wikipedia page when you are logged in), and then selecting "Image" from the dropdown box. Note that any non-free media lacking such an explanation will be deleted one week after they have been uploaded, as described on criteria for speedy deletion. If you have any questions please ask them at the Media copyright questions page. Thank you. Ricky81682 (talk) 08:28, 26 October 2007 (UTC)

Bookmark

http://maxima.sourceforge.net/docs/intromax/intromax.html

http://maxima.sourceforge.net/docs/tutorial/en/gaertner-tutorial-revision/Contents.htm

http://beshenov.ru/maxima/faq.html

http://math.nist.gov/~BMiller/computer-algebra/

http://www.delorie.com/gnu/docs/maxima/maxima_toc.html#SEC_Contents

http://maxima.sourceforge.net/docs/manual/en/maxima.html

math

$\left[z=e^{{i\,\pi } \over {2}},z=e^{{5\,i\,\pi } \over {6}},z=e^{{7\,i\,\pi } \over {6}},z=e^{-{{i\,\pi } \over {2}}},z=e^{-{{i\,\pi } \over {6}}},z=e^{{i\,\pi } \over {6}}\right]$

License tagging for File:Obama wining 20121104.png

Thanks for uploading File:Obama wining 20121104.png. You don't seem to have indicated the license status of the image. Wikipedia uses a set of image copyright tags to indicate this information.

To add a tag to the image, select the appropriate tag from this list, click on this link, then click "Edit this page" and add the tag to the image's description. If there doesn't seem to be a suitable tag, the image is probably not appropriate for use on Wikipedia. For help in choosing the correct tag, or for any other questions, leave a message on Wikipedia:Media copyright questions. Thank you for your cooperation. --ImageTaggingBot (talk) 03:05, 4 November 2012 (UTC)

Integration by substitution

The objective of Integration by substitution is to substitute the integrand with $u$ or $g(x)$

Theory

We want to transform the Integral from a function of x to a function of u

$\int _{x=a}^{x=b}f(x)\,dx\,\rightarrow \,\int _{u=c}^{u=d}h(u)\,du$

Starting with

$u\,=\,g(x)$
and

$\int _{x=a}^{x=b}f(x)\,dx\,$	$=\int _{x=a}^{x=b}f(x)\,{\operatorname {d} \!u \over \operatorname {d} \!u}\,dx\,$		(1)	ie   ${\operatorname {d} \!u \over \operatorname {d} \!u}\,=\,1$
	$=\int _{x=a}^{x=b}(f(x)\,{\operatorname {d} \!x \over \operatorname {d} \!u})({\operatorname {d} \!u \over \operatorname {d} \!x})\,dx\,$		(2)	ie   ${\operatorname {d} \!x \over \operatorname {d} \!u}{\operatorname {d} \!u \over \operatorname {d} \!x}\,=\,{\operatorname {d} \!u \over \operatorname {d} \!u}\,=\,1$
	$=\int _{x=a}^{x=b}(f(x)\,{\operatorname {d} \!x \over \operatorname {d} \!u})g'(x)\,dx\,$		(3)	ie   ${\operatorname {d} \!u \over \operatorname {d} \!x}\,=\,g'(x)$
	$=\int _{x=a}^{x=b}h(g(x))g'(x)\,dx\,$		(4)	ie   Now equate $(f(x)\,{\operatorname {d} \!x \over \operatorname {d} \!u})$ with $h(g(x))$
	$=\int _{x=a}^{x=b}h(u)g'(x)\,dx\,$		(5)	ie   $g(x)\,=\,u$
	$=\int _{u=g(a)}^{u=g(b)}h(u)\,du\,$		(6)	ie   $du\,=\,{\operatorname {d} \!u \over \operatorname {d} \!x}dx\,=\,g'(x)\,dx\,$
	$=\int _{u=c}^{u=d}h(u)\,du\,$		(7)	ie   We have achieved our desired result

Procedure

• Calculate $g'(x)\,=\,{\operatorname {d} \!u \over \operatorname {d} \!x}$
• Calculate $h(u)$ which is $f(x)\,{\operatorname {d} \!x \over \operatorname {d} \!u}\,=\,{\frac {f(x)}{g'(x)}}$ and make sure you express the result in terms of the variable u
• Calculate $c\,=\,g(a)$
• Calculate $d\,=\,g(b)$

ArbCom elections are now open!

Hi,
You appear to be eligible to vote in the current Arbitration Committee election. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to enact binding solutions for disputes between editors, primarily related to serious behavioural issues that the community has been unable to resolve. This includes the ability to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate, you are welcome to review the candidates' statements and submit your choices on the voting page. For the Election committee, MediaWiki message delivery (talk) 13:00, 23 November 2015 (UTC)

ArbCom Elections 2016: Voting now open!

Hello, Ohanian. Voting in the 2016 Arbitration Committee elections is open from Monday, 00:00, 21 November through Sunday, 23:59, 4 December to all unblocked users who have registered an account before Wednesday, 00:00, 28 October 2016 and have made at least 150 mainspace edits before Sunday, 00:00, 1 November 2016.

The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail.

If you wish to participate in the 2016 election, please review the candidates' statements and submit your choices on the voting page. MediaWiki message delivery (talk) 22:08, 21 November 2016 (UTC)

ArbCom 2017 election voter message

Hello, Ohanian. Voting in the 2017 Arbitration Committee elections is now open until 23.59 on Sunday, 10 December. All users who registered an account before Saturday, 28 October 2017, made at least 150 mainspace edits before Wednesday, 1 November 2017 and are not currently blocked are eligible to vote. Users with alternate accounts may only vote once.

The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail.

If you wish to participate in the 2017 election, please review the candidates and submit your choices on the voting page. MediaWiki message delivery (talk) 18:42, 3 December 2017 (UTC)

What is the minimum amount of information to rotate a point in 3D Cartesian space?

Suppose you are given a 3D cartesian point {x,y,z} = {1,2,3}. What is the minimum information you need to rotate that point in 3D space to another point in 3D space? For many years I thought all I need is

1. X coordinate of rotation origin
2. Y coordinate of rotation origin
3. Z coordinate of rotation origin
4. X coordinate of the normal vector
5. Y coordinate of the normal vector
6. Z coordinate of the normal vector
7. Angle of rotation in either degrees or radians
8. Rotation using the normal vector using which rule? Right Hand Rule or Left Hand Rule?

But I was mistaken, because I got the wrong answers to my problem. It took me a few days until I realized what my problem is.

I am missing the Cartesian Axis Type.

For example:

Cartesian Axis Type 1: X-axis is towards the observer, Y-axis is to the right (from observer's point of view), Z-axis is upwards (from observer's point of view)

Cartesian Axis Type 2: X-axis is to the right (from observer's point of view), Y-axis is upwards (from observer's point of view), Z-axis is away from the observer

Depending on which Cartesian Axis Type, you will get completely different answers.

So what is the minimum amount of information to rotate a point in 3D Cartesian space? Am I missing anything else? Ohanian (talk) 03:37, 30 December 2017 (UTC)

Why is it so hard to solve this puzzle in other programming languagues?

Why is it so hard to solve this puzzle in other programming languages like basic, C, perl or python?

Here is the puzzle, it is from the url youtu.be/RY7YKSw1t_M

There are 6 apple seeds. Everyday, there is 1/2 chance that each seed will turn into an apple tree. Once a seed turn into an apple tree, it will remain an apple tree forever. What is the expected number of days when all six seeds has finally turn into apple trees.

The answer is 7880/1953 which is calculated using wolfram/mathematica language source below.

h[0]=0;
func[n_]:=Module[
  {expr,soln},
  expr = h[n] == Sum[Binomial[n, k]*(1/2)^n*(1 + h[n - k]), {k, 0, n}];
  soln = Part[  Solve[expr,h[n]], 1  ];
  h[n] = h[n] /. soln;
  h[n]
];
Map[func, {1,2,3,4,5,6} ]

With the result

{2,8/3,22/7,368/105,2470/651,7880/1953}

How do I write the program above in python? Why is it so difficult to write the program above in any other programming languages? Ohanian (talk) 16:55, 25 January 2018 (UTC)

Estimating Pi using Zeta Function

We have the identity of

${\frac {\pi ^{2}}{6}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots$

which can be used to compute the value of $\pi$ but it very very slowly because of the denominator terms which only increases in the second power. Other zeta function identities are

${\frac {\pi ^{4}}{90}}={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+\cdots$

${\frac {\pi ^{6}}{945}}={\frac {1}{1^{6}}}+{\frac {1}{2^{6}}}+{\frac {1}{3^{6}}}+\cdots$

But we shall use the below because of the numeric value of 10 as it will make the calculations easier when we uses the common log function.

${\frac {\pi ^{10}}{93555}}={\frac {1}{1^{10}}}+{\frac {1}{2^{10}}}+{\frac {1}{3^{10}}}+\cdots$

Let's simplify by defining Y as

$Y={\frac {1}{1^{10}}}+{\frac {1}{2^{10}}}+{\frac {1}{3^{10}}}+\cdots$

Thus

${\frac {\pi ^{10}}{93555}}=Y$

$\pi ^{10}=93555\times Y$

$\log _{10}(\pi ^{10})=\log _{10}(93555)+\log _{10}(Y)$

$10\times \log _{10}(\pi )=\log _{10}(93555)+\log _{10}(Y)$

$\log _{10}(\pi )={\frac {1}{10}}\times \log _{10}(93555)+{\frac {1}{10}}\times \log _{10}(Y)$

$\pi =10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times \log _{10}(Y)}$

At this point we shall pull a trick be defining an approximation of $Y$ called $Y_{n}$

$Y_{n}={\frac {1}{1^{10}}}+{\frac {1}{2^{10}}}+{\frac {1}{3^{10}}}+\cdots +{\frac {1}{n^{10}}}=\sum _{k=1}^{n}{\frac {1}{k^{10}}}$

$Y_{n}=Y_{n-1}+{\frac {1}{n^{10}}}$

$Y=Y_{\infty }$

Pay special attention to $Y_{1}$ it would come useful later

$Y_{1}={\frac {1}{1^{10}}}=1$

To recap we have

$\pi =10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times \log _{10}(Y_{\infty })}$

At this point we shall pull another trick be defining an approximation of $\pi$ called $\pi _{n}$

$\pi _{n}=10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times \log _{10}(Y_{n})}$

$\lim _{n\to \infty }\pi _{n}=\pi _{\infty }=\pi$

Thus

$\pi =\pi _{\infty }$

Pay special attention to $\pi _{1}$

$\pi _{1}=10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times \log _{10}(Y_{1})}$

$\pi _{1}=10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times \log _{10}(1)}$

$\pi _{1}=10^{{\frac {1}{10}}\times \log _{10}(93555)}$

Now we pull yet another trick

$log_{10}(A+B)=log_{10}(A)+log_{10}(1+{\frac {B}{A}})$

Thus

$log_{10}(Y_{n})=log_{10}(Y_{n-1}+{\frac {1}{n^{10}}})=log_{10}(Y_{n-1})+log_{10}(1+{\frac {\frac {1}{n^{10}}}{Y_{n-1}}})$

And finally we have our result

$\pi _{n}=10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times (log_{10}(Y_{n-1})+log_{10}(1+{\frac {\frac {1}{n^{10}}}{Y_{n-1}}}))}$

We simplify

$\pi _{n}=10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times (log_{10}(Y_{n-1})}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{n^{10}}}{Y_{n-1}}}))}$

Next we replace $10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times (log_{10}(Y_{n-1})}$ with $\pi _{n-1}$

$\pi _{n}=\pi _{n-1}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{n^{10}}}{Y_{n-1}}}))}$

Summary, to calculate the value of $\pi$, we first

(1) Calc $Y_{1}={\frac {1}{1^{10}}}=1$

(2) Calc $\pi _{1}=10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times \log _{10}(Y_{1})}=10^{{\frac {1}{10}}\times \log _{10}(93555)}$

(3) Calc $\pi _{2}=\pi _{1}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{2^{10}}}{Y_{1}}}))}$

(4) Calc $Y_{2}=Y_{1}+{\frac {1}{2^{10}}}$

(5) Repeat step 3 and 4 for $\pi _{n}$ and $Y_{n}$ using

$\pi _{n}=\pi _{n-1}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{n^{10}}}{Y_{n-1}}}))}$

and

$Y_{n}=Y_{n-1}+{\frac {1}{n^{10}}}$

There is a problem of course, the problem is that the calculations are done on a calculator. For example my high school scientific calculator can only display 10 significant digits, which means the smallest number greater than 1 that the calculator can display is 1.000000001 and this means that

$1+{\frac {\frac {1}{n^{10}}}{Y_{n-1}}}>1.000000001$ for a 10 digits calculator

We know that $Y_{n-1}\approx 1$ so we can approximate and simplify to

$1+{\frac {1}{n^{10}}}>1.000000001$

${\frac {1}{n^{10}}}>0.000000001$

${\frac {1}{n^{10}}}>1\times 10^{-9}$

$n^{10}<1\times 10^{9}$

$n<{\sqrt[{10}]{1\times 10^{9}}}$

$n<7.94$

So we can only calculator the first 7 terms from n = 1 to 7

It may be more accurate to calculate directly

$\pi =10^{{\frac {1}{10}}\times \log _{10}(93555)}\times 10^{{\frac {1}{10}}\times \log _{10}(Y)}$

But $Y$ is just the sum of various ${\frac {1}{n^{10}}}$ which means that it must be greater than the smallest number greater than zero for that particular calculator.

${\frac {1}{n^{10}}}>1\times 10^{-99}$

$n^{10}<1\times 10^{99}$

$n<7943282347$

So you may think that n can go all the way to a huge number but

$Y_{n}=1+{\frac {1}{2^{10}}}+{\frac {1}{3^{10}}}+\cdots +{\frac {1}{n^{10}}}$

So for the last term ${\frac {1}{n^{10}}}$ to make any practical difference to $Y_{n}$ we have the same problem as before!

${\frac {1}{n^{10}}}>0.000000001$

${\frac {1}{n^{10}}}>1\times 10^{-9}$

$n^{10}<1\times 10^{9}$

$n<{\sqrt[{10}]{1\times 10^{9}}}$

$n<7.94$

Using my high school scientific calculator with 10 significant digits, I get

$\pi _{1}=10^{{\frac {1}{10}}\times \log _{10}(93555)}=3.141280369$

$\pi _{2}=\pi _{1}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{2^{10}}}{Y_{1}}}))}$

$\pi _{2}=\pi _{1}\times 1.000097613=3.141587000$

$Y_{2}=Y_{1}+{\frac {1}{2^{10}}}=1+9.765625000\times 10^{-4}=1.000976563$

$\pi _{3}=\pi _{2}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{3^{10}}}{Y_{2}}}))}$

$\pi _{3}=3.141587000\times 1.000001692=3.141592315$

$Y_{3}=Y_{2}+{\frac {1}{3^{10}}}=1.000976563+1.693508781\times 10^{-5}=1.000993498$

$\pi _{4}=\pi _{3}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{4^{10}}}{Y_{3}}}))}$

$\pi _{4}=3.141592315\times 1.000000095=3.141592614$

$Y_{4}=Y_{3}+{\frac {1}{4^{10}}}=1.000993498+9.536743164\times 10^{-7}=1.000994452$

$\pi _{5}=\pi _{4}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{5^{10}}}{Y_{4}}}))}$

$\pi _{5}=3.141592614\times 1.000000010=3.141592646$

$Y_{5}=Y_{4}+{\frac {1}{5^{10}}}=1.000994452+1.024000000\times 10^{-7}=1.000994554$

$\pi _{6}=\pi _{5}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{6^{10}}}{Y_{5}}}))}$

$\pi _{6}=3.141592646\times 1.000000002=3.141592651$

$Y_{6}=Y_{5}+{\frac {1}{6^{10}}}=1.000994554+1.653817169\times 10^{-8}=1.000994571$

$\pi _{7}=\pi _{6}\times 10^{{\frac {1}{10}}\times log_{10}(1+{\frac {\frac {1}{7^{10}}}{Y_{6}}}))}$

$\pi _{7}=3.141592651\times 1.000000000=3.141592651$

Let's calculate $\pi$ directly

$Y_{7}=\sum _{k=1}^{7}{\frac {1}{k^{10}}}=1.000994574$

$\pi _{7}=\pi _{1}\times 10^{{\frac {1}{10}}\times \log _{10}(Y_{7})}$

$\pi _{7}=3.141280369\times 1.000099413=3.141592653$

Estimating Pi using Zeta Function using n^18

We have the identity of

${\frac {\pi ^{2}}{6}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots$

which can be used to compute the value of $\pi$ but it very very slowly because of the denominator terms which only increases in the second power. Other zeta function identities are

${\frac {\pi ^{4}}{90}}={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+\cdots$

${\frac {\pi ^{6}}{945}}={\frac {1}{1^{6}}}+{\frac {1}{2^{6}}}+{\frac {1}{3^{6}}}+\cdots$

But we shall use the below because the denominator have integers to the 18th power which will grow large very very fast indeed.

${\frac {43867\pi ^{18}}{38979295480125}}={\frac {1}{1^{18}}}+{\frac {1}{2^{18}}}+{\frac {1}{3^{18}}}+\cdots$

Let's simplify by defining Y as

$Y={\frac {1}{1^{18}}}+{\frac {1}{2^{18}}}+{\frac {1}{3^{18}}}+\cdots$

Thus

${\frac {43867\pi ^{18}}{38979295480125}}=Y$

$\pi ^{18}={\frac {38979295480125}{43867}}\times Y$

$\log _{10}(\pi ^{18})=\log _{10}({\frac {38979295480125}{43867}})+\log _{10}(Y)$

$18\times \log _{10}(\pi )=\log _{10}({\frac {38979295480125}{43867}})+\log _{10}(Y)$

$\log _{10}(\pi )={\frac {1}{18}}\times \log _{10}({\frac {38979295480125}{43867}})+{\frac {1}{18}}\times \log _{10}(Y)$

$\pi =10^{{\frac {1}{18}}\times \log _{10}({\frac {38979295480125}{43867}})}\times 10^{{\frac {1}{18}}\times \log _{10}(Y)}$

At this point we shall pull a trick be defining an approximation of $Y$ called $Y_{n}$

$Y_{n}={\frac {1}{1^{18}}}+{\frac {1}{2^{18}}}+{\frac {1}{3^{18}}}+\cdots +{\frac {1}{n^{18}}}=\sum _{k=1}^{n}{\frac {1}{k^{18}}}$

$Y=Y_{\infty }$

To recap we have

$\pi =10^{{\frac {1}{10}}\times \log _{10}({\frac {38979295480125}{43867}})}\times 10^{{\frac {1}{10}}\times \log _{10}(Y_{\infty })}$

At this point we shall pull another trick be defining an approximation of $\pi$ called $\pi _{n}$

$\pi _{n}=10^{{\frac {1}{18}}\times \log _{10}({\frac {38979295480125}{43867}})}\times 10^{{\frac {1}{18}}\times \log _{10}(Y_{n})}$

$\lim _{n\to \infty }\pi _{n}=\pi _{\infty }=\pi$

Thus

$\pi =\pi _{\infty }$

Pay special attention to $\pi _{1}$

$\pi _{1}=10^{{\frac {1}{18}}\times \log _{10}({\frac {38979295480125}{43867}})}\times 10^{{\frac {1}{18}}\times \log _{10}(Y_{1})}$

$\pi _{1}=10^{{\frac {1}{18}}\times \log _{10}({\frac {38979295480125}{43867}})}\times 10^{{\frac {1}{18}}\times \log _{10}(1)}$

$\pi _{1}=10^{{\frac {1}{18}}\times \log _{10}({\frac {38979295480125}{43867}})}$

$\pi _{n}=\pi _{1}\times 10^{{\frac {1}{18}}\times \log _{10}(Y_{n})}$

There is a problem of course, the problem is that the calculations are done on a calculator. For example my high school scientific calculator can only display 10 significant digits, which means the smallest number greater than 1 that the calculator can display is 1.000000001 and this means that for the last term ${\frac {1}{n^{18}}}$ to make any practical difference to $Y_{n}$ we have the problem that $1+{\frac {1}{n^{18}}}>1+0.000000001$

Given $Y_{n}=1+{\frac {1}{2^{18}}}+{\frac {1}{3^{18}}}+\cdots +{\frac {1}{n^{18}}}$ and for the last term to have a difference on a 10 digits calculator

${\frac {1}{n^{18}}}>0.000000001$

${\frac {1}{n^{18}}}>1\times 10^{-9}$

$n^{18}<1\times 10^{9}$

$n<{\sqrt[{18}]{1\times 10^{9}}}$

$n<3.16$

Let's calculate $\pi$ directly using only the first 3 terms on $Y$ thus n=3

$Y_{3}=\sum _{k=1}^{3}{\frac {1}{k^{18}}}={\frac {1}{1^{18}}}+{\frac {1}{2^{18}}}+{\frac {1}{3^{18}}}=1.000003817$

$\pi _{1}=10^{{\frac {1}{18}}\times \log _{10}({\frac {38979295480125}{43867}})}=3.141591987$

$\pi _{3}=\pi _{1}\times 10^{{\frac {1}{18}}\times \log _{10}(Y_{3})}$

$\pi _{3}=3.141591987\times 1.000000212=3.141592653$

ArbCom 2018 election voter message

Hello, Ohanian. Voting in the 2018 Arbitration Committee elections is now open until 23.59 on Sunday, 3 December. All users who registered an account before Sunday, 28 October 2018, made at least 150 mainspace edits before Thursday, 1 November 2018 and are not currently blocked are eligible to vote. Users with alternate accounts may only vote once.

The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail.

If you wish to participate in the 2018 election, please review the candidates and submit your choices on the voting page. MediaWiki message delivery (talk) 18:42, 19 November 2018 (UTC)

How many digits of pi for the known universe?

If I wanted to measure the circumference of the known universe to the accuracy of a planck length, how many digits of pi should I use?

PS. I'm assuming that the universe can be modelled by a very large circle.

202.168.50.40 02:47, 27 September 2006 (UTC)

It's impossible to make such an precise measurement, and even if you could, it would take a sophisticated experimental setup, not any number of digits of some mathematical constant. Melchoir 02:54, 27 September 2006 (UTC)

That's a pretty important assumption- see Shape of the universe. At very small and very large distances, we tend to have trouble modelling shapes and distances, and the universe almost certainly can't be modelled by a simple circle. --froth^{T C} 02:56, 27 September 2006 (UTC)

all of them. anything less simply won't do. Xcomradex 03:18, 27 September 2006 (UTC)

The original question seems pretty reasonable to me. According to Observable universe#size, our universe, modeled as a sphere, is between 14 and 78 billion light-years in radius. I'm not going to try to figure out the answer, at least not tonight, but it should be doable. --Allen 03:38, 27 September 2006 (UTC)

I'm assuming, though, that the questioner is really talking about calculating the circumference of the universe, not measuring it... so that it's a math problem, not a physics problem. --Allen 03:43, 27 September 2006 (UTC)

With that data, it's easy: you have at most one significant digit to work with, so "pi = 3." suffices. It won't get you Planck-level accuracy, but neither will any other method without better data. Melchoir 04:21, 27 September 2006 (UTC)

Please see Pi#Numerical value for a start to answering the question. — Knowledge Seeker দ 04:13, 27 September 2006 (UTC)

Your error is going to be far larger than one Planck length, unless you make the assumption from the start that the measurements you're using are perfectly accurate. -- Consumed Crustacean (talk) 04:16, 27 September 2006 (UTC)

Radius = 78 billion light years = 78E9 * 9.461E15 = 7.38E26 metres

Diameter = D = 2 * 7.38E26 metres = 1.476E27 metres

Let C be the circumference

$C=\pi *D\,$

thus

$C+\Delta C=(\pi +\Delta \pi )*D\,$

$C+\Delta C=\pi *D+\Delta \pi *D\,$

$\Delta C=\Delta \pi *D\,$

$\Delta \pi ={\frac {\Delta C}{D}}\,$

$\Delta \pi ={\frac {\mbox{planck length}}{D}}\,$

$\Delta \pi ={\frac {1.6*10^{-35}}{1.476*10^{27}}}\,$

$\Delta \pi =1.08*10^{-62}\,$

So I say you need about 63 or 64 digits of PI.

I still agree with the anon calculator, because I think the original poster meant to ask about the number of digits of pi given a known radius of the universe. And as long as the universe's radius is in that tens-of-billions range, the answer is going to be 63 or 64 digits. And it's an interesting question, too... insofar as using pi to calculate circumferences is concerned, we will never ever need more than 64 digits for any practical problem. I bet that's what led the questioner to ask the question. --Allen 04:30, 27 September 2006 (UTC)

The point of the question is that we long ago calculated pi to far more digits than would be needed to calculate the diameter/circumference/radius of the observable universe to subatomic uncertainties. The had it to 100,000 decimals by 1968, if I recall correctly. I believe Isaac Asimov pointed this out several decades ago. So going from 100 digits to 100,000, to 100,000,000 perhaps does not have any practical purpose. Edison 05:26, 27 September 2006 (UTC)

pi's used for a lot more things than calculating circumferences. In which one of his 300 books did asimov say this?

Perhaps "Asimov on Numbers" but I read it decades ago.Edison 14:48, 27 September 2006 (UTC)

ArbCom 2024 Elections voter message

Hello! Voting in the 2024 Arbitration Committee elections is now open until 23:59 (UTC) on Monday, 2 December 2024. All eligible users are allowed to vote. Users with alternate accounts may only vote once.

The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail.

If you wish to participate in the 2024 election, please review the candidates and submit your choices on the voting page. If you no longer wish to receive these messages, you may add {{NoACEMM}} to your user talk page. MediaWiki message delivery (talk) 00:09, 19 November 2024 (UTC)